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$\int \frac{d x}{x \sqrt{x^{6}-16}}$
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Verified Answer
The correct answer is:
$\frac{1}{12} \sec ^{-1}\left(\frac{x^{3}}{4}\right)+c$
Let $I=\int \frac{d x}{x \sqrt{\left(x^{3}\right)^{2}-16}}$
Put $x^{3}=t \Rightarrow 3 x^{2} d x=d t$
So, $I=\frac{1}{3} \int \frac{d t}{x^{3} \sqrt{\left(x^{3}\right)^{2}-16}}$
$\begin{aligned}
&=\frac{1}{3} \int \frac{d t}{t\left(\sqrt{t^{2}-16}\right.} \\
&=\frac{1}{3 \times 4} \sec ^{-1}\left(\frac{t}{4}\right)+c \\
&=\frac{1}{12} \sec ^{-1}\left(\frac{x^{3}}{4}\right)+c
\end{aligned}$
Put $x^{3}=t \Rightarrow 3 x^{2} d x=d t$
So, $I=\frac{1}{3} \int \frac{d t}{x^{3} \sqrt{\left(x^{3}\right)^{2}-16}}$
$\begin{aligned}
&=\frac{1}{3} \int \frac{d t}{t\left(\sqrt{t^{2}-16}\right.} \\
&=\frac{1}{3 \times 4} \sec ^{-1}\left(\frac{t}{4}\right)+c \\
&=\frac{1}{12} \sec ^{-1}\left(\frac{x^{3}}{4}\right)+c
\end{aligned}$
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